Why does this mathematical series plateau below $\frac{1}{4}$ then have runaway growth? This recursive mathematical series plateaus when $x \leq 1/4$, and it then goes runaway growth. Why?
x = 1/4
y = 0
for i in range(10000000):
    y = (x + y)**2

 A: As a remark, you seem to implicitely only allow non-negative $x$ (setting say $x=-1000$ certainly leads to a runaway growth of $y$), so I'll keep that restriction.
So why is there no runaway growth for $0 \le x \le \frac14$?
As Matti.P wrote in a comment, your loop code would mathematically be described as a seqeunce: $y_0=0;\; y_{i+1}=(x+y_i)^2$ for all $i=0,1,2,\ldots$.
You start with $y_0 = 0 \le \frac14$.
Whenever you do the next loop iteration, and start with an $y_i$ that fulfills $0 \le y_i \le \frac14$, then the next $y_{i+1}$ fulfills the same condition!
Let's prove it. From
$$ 0 \le x+y_i \le \frac14 + \frac14 =\frac12,$$
which comes from using $0 \le x \le \frac14$ and $0 \le y_i \le \frac14$, we can immediately conclude, as the function $f(t)=t^2$ is increasing for $0 \le t$, that
$$ 0^2=0 \le y_{i+1}=(x+y_i)^2 \le \frac14 = \left(\frac12\right)^2.$$
So if $0 \le x \le \frac14$ and because we start with an $y$ value that is between $0$ and $\frac14$, all the following $y$-values will also stay in that interval.
To see why there is runaway growth for $x > \frac14$, we need a little bit more theory.
First, for all $x>0$ the sequence of $(y_i)$ will be increasing. It's true for the very first step from $y_0$ to $y_1$:
$$y_0=0; y_1=x^2 > 0.$$
And it keeps true from one step to the next:
if $0 \le y_i < y_{i+1}$, then
$$y_{i+1}=(x+y_i)^2 < (x+y_{i+1})^2 = y_{i+2},$$
again using the (strict) monotonicity of $f(t)=t^2$ for $t \ge 0$.
So $(y_n)$ is increasing for $x>0$. Such sequences can only have 2 behaviours:

*

*they converge to a limit, or

*they increase beyond all bounds and "converge" to $+\infty$.

We've seen above that for $0 \le x \le \frac14$ there is no increase beyond all bounds, so it must converge to a limit $l(x)$. How can this limit be calculated?
Well if we know (or assume) that $\lim_{i\to\infty}y_i=l(x)$, then we know that
$$l(x)=\lim_{i\to\infty}y_i = \lim_{i\to\infty}y_{i+1} = \lim_{i\to\infty} (x+y_i)^2 =(x+l(x))^2,$$
where the last equality used the fact that the function $g(t)=(x+t)^2$ is continous ($x$ is just a constant for that function). As we can see, this yields a quadratic equation for $l(x)$, which you can solve the usual way and get
$$l(x)=\frac{1-2x}2 \pm \sqrt{\frac{1-4x}4}$$
Which of the 2 values is actually the limit depends on $y_0$, but this is not our concern here. Note that the term under the square root is $\frac{1-4x}4$. That means the square root exists (in real numbers) only when $x \le \frac14$. If $x >\frac14$, the square root is complex and the only solutions to the equation are 2 non-real numbers.
But obviously, our sequence contains only real numbers, so the limit (if it exists) must be a real number. So the only conclusion we can draw is that for $x > \frac14$, the sequence $(y_n)$ has no limit.
But as I wrote above, for increasing sequences there are only 2 kinds of behaviours, and one has just been ruled out. So it has to be the other, which is unbounded growth.
